Skip to main content
Chemistry LibreTexts

14.15: Problems

  • Page ID
    152683
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Problems

    1. When we express the energy of a system as a function of entropy, volume, and composition, we have \(E=E\left(S,V,n_1,n_2,\dots ,\ n_{\omega }\right)\). Since \(S\) and \(V\) are extensive variables, we have \(\lambda E=E\left(\lambda S,\lambda V,{\lambda n}_1,\lambda n_2,\dots ,\ \lambda n_{\omega }\right)\). Find \({\left({\partial \left(\lambda E\right)}/{\partial \lambda }\right)}_{SV}\). From this result, show that \[G=\sum^{\omega }_{j=1}{{\mu }_jn_j} \nonumber \]

    2. When we express the energy of a system as a function of pressure, temperature, and composition, we have \(E=E\left(P,T,n_1,n_2,\dots ,\ n_{\omega }\right)\). Because P and T are independent of \(\lambda\), \(\lambda E=E\left(P,T,{\lambda n}_1,\lambda n_2,\dots ,\ \lambda n_{\omega }\right)\). Show that

    \[E=\sum^{\omega }_{j=1} \overline{E}_jn_j \nonumber \]

    3. From \(E\mathrm{=E}\left(P,T,n_{\mathrm{1}},n_{\mathrm{2}}\mathrm{,\dots ,\ }n_{\omega }\right)\) and the result in problem 2, show that

    \[\left[{\left(\frac{\partial H}{\partial T}\right)}_P + P{\left(\frac{\partial V}{\partial T}\right)}_P\right]dT + \left[P{\left(\frac{\partial V}{\partial P}\right)}_T + T{\left(\frac{\partial V}{\partial T}\right)}_P\right]dP = \sum^{\omega }_{j\mathrm{=1}}{n_j}d{\overline{E}}_j \nonumber \]

    Note that at constant pressure and temperature,

    \[\sum^{\omega }_{j\mathrm{=1}}{n_j}d{\overline{E}}_j\mathrm{=0} \nonumber \]

    4. If pressure and temperature are constant, \(E=E\left(n_1,n_2,\dots ,\ n_{\omega }\right)\) and \(\lambda E=E\left({\lambda n}_1,\lambda n_2,\dots ,\ \lambda n_{\omega }\right)\). Show that \(\sum^{\omega }_{j\mathrm{=1}}{n_j}d{\overline{E}}_j\mathrm{=0}\) follows from these relationships.

    5. A solution contains \(n_1\) moles of component 1, \(n_2\) moles of component 2, \(n_3\) moles of component 3, etc. Let \(n=n_1+n_2+n_3+...\) The mole fraction of component \(j\) is \(x_j={n_j}/{n}\). Show that \[\left(\frac{\partial x_j}{\partial n_j}\right)=\frac{n-n_j}{n^2} \nonumber \] and, for \(j\neq k\), \[\ \left(\frac{\partial x_j}{\partial n_k}\right)=\frac{-n_j}{n^2} \nonumber \] What are \[\left(\frac{\partial x_1}{\partial n_1}\right) \nonumber \] and \[\left(\frac{\partial x_2}{\partial n_2}\right) \nonumber \] if the solution has only two components?

    6. For any extensive state function, \(Y\left(P,T,n_1,n_2,\dots ,\ n_{\omega }\right)\), the arguments developed in this chapter lead, at constant \(P\) and\(\ T\), to the equations

    \[Y=n_1{\overline{Y}}_1+n_2{\overline{Y}}_2+\dots +n_{\omega }{\overline{Y}}_{\omega } \nonumber \] and \[0=n_1d{\overline{Y}}_1+n_2d{\overline{Y}}_2+\dots +n_{\omega }d{\overline{Y}}_{\omega } \nonumber \]

    Where \({\overline{Y}}_j\) is the partial molar quantity \({\left({\partial Y}/{\partial n_j}\right)}_{P,T,n_{m\neq j}}\).

    (a) Prove that \(0=x_1d{\overline{Y}}_1+x_2d{\overline{Y}}_2+\dots +x_{\omega }d{\overline{Y}}_{\omega }\)

    (b) Prove that \[0=n_1\left(\frac{\partial {\overline{Y}}_1}{\partial n_1}\right)+n_2\left(\frac{\partial {\overline{Y}}_2}{\partial n_2}\right)+\dots +n_{\omega }\left(\frac{\partial {\overline{Y}}_{\omega }}{\partial n_{\omega }}\right) \nonumber \] (c) Prove that \[0=x_1\left(\frac{\partial {\overline{Y}}_1}{\partial x_1}\right)+x_2\left(\frac{\partial {\overline{Y}}_2}{\partial x_2}\right)+\dots +x_{\omega }\left(\frac{\partial {\overline{Y}}_{\omega }}{\partial x_{\omega }}\right) \nonumber \]

    7. The enthalpy of mixing is measured in a series of experiments in which solid solute, \(A\), dissolves to form an aqueous solution. These enthalpy data are represented well by empirical equations \({\Delta }_{mix}H={\alpha }_1\underline{m}+{\alpha }_2{\underline{m}}^2\), \({\alpha }_1={\beta }_{11}+{\beta }_{12}\left(T-273.15\right)\) and

    \({\alpha }_2={\beta }_{21}+{\beta }_{22}\left(T-273.15\right)\) with \[{\beta }_{11}=10.0\ \mathrm{kJ}\ {\mathrm{molal}}^{-1} \nonumber \] \[{\beta }_{12}=-0.14\ \mathrm{kJ}\ {\mathrm{molal}}^{-2}\ K^{-1} \nonumber \] \[{\beta }_{21}=-3.00\ \mathrm{kJ}\ {\mathrm{molal}}^{-1} \nonumber \] \[{\beta }_{22}=-0.040\ \mathrm{kJ}\ {\mathrm{molal}}^{-2}\ K^{-1} \nonumber \] Find \({\overline{L}}_A\), \({\overline{L}}_{H_2O}\), \({\overline{J}}_A\), and \({\overline{J}}_{H_2O}\) as functions of \({\underline{m}}_A\) and \(T\). Find \({\overline{L}}_A\), \({\overline{L}}_{H_2O}\), \({\overline{J}}_A\), and \({\overline{J}}_{H_2O}\) for a one molal solution at 209 K. What is the value of

    \[{ \ln \frac{{\tilde{a}}_A\left(1\mathrm{\ molal},290\mathrm{\ K}\right)}{{\tilde{a}}_A\left(1\mathrm{\ molal},273.15\mathrm{\ K}\right)}\ } \nonumber \]

    Notes

    \({}^{1}\) We can make other assumptions. It is possible to describe an inhomogeneous system as a collection of many macroscopic, approximately homogeneous regions.


    This page titled 14.15: Problems is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to the style and standards of the LibreTexts platform.